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Step-by-Step Solution
Step 1: Identify the Sample Space
When a fair die is thrown twice, there are 36 equally likely outcomes
(each outcome is an ordered pair $(x, y)$ where $x$ and $y$ can each be from $1$ to $6$).
Step 2: Define the Event A (Sum is a Multiple of 4)
Let $A$ be the event that the sum of the two throws is a multiple of 4.
The possible sums that are multiples of 4 from two dice are 4, 8, and 12.
Sum $= 4$: $(1, 3), (2, 2), (3, 1)$
Sum $= 8$: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$
Sum $= 12$: $(6, 6)$
Therefore, $A = \{(1,3), (3,1), (2,2), (2,6), (6,2), (3,5), (5,3), (4,4), (6,6)\}.$
There are $9$ such outcomes in $A$, hence $n(A) = 9.$
Step 3: Define the Event B (Score 4 Appears at Least Once)
Let $B$ be the event that the outcome shows a 4 on at least one of the dice.
In general, $B$ would include pairs such as $(4, 1), (1, 4), (4, 2), (2, 4), \dots$
but since we are dealing with the conditional probability given that we are already in event $A,$
we only look for outcomes in $A$ that contain at least one 4.
From the list of outcomes in $A,$ the only pair that includes a 4 is $(4,4).$
So $A \cap B = \{(4,4)\}.$
Step 4: Compute the Required Conditional Probability
We want $P(B \mid A)$, which is given by
$$
P(B \mid A) = \frac{P(A \cap B)}{P(A)}.
$$
Since each of the 36 outcomes is equally likely,
$$
P(A) = \frac{n(A)}{36} = \frac{9}{36} = \frac{1}{4},
$$
and
$$
P(A \cap B) = \frac{n(A \cap B)}{36} = \frac{1}{36}.
$$
Therefore,
$$
P(B \mid A) = \frac{\frac{1}{36}}{\frac{9}{36}} = \frac{1}{9}.
$$
Step 5: State the Final Answer
Hence, the conditional probability that a 4 appears at least once,
given that the sum is a multiple of 4, is
$$
\frac{1}{9}.
$$
